Problem: $ABCDEFGH$ shown below is a cube. Find $\sin \angle HAC$.

[asy]

import three;

triple A,B,C,D,EE,F,G,H;

A = (0,0,0);

B = (1,0,0);

C = (1,1,0);

D= (0,1,0);

EE = (0,0,1);

F = B+EE;

G = C + EE;

H = D + EE;

draw(B--C--D);

draw(B--A--D,dashed);

draw(EE--F--G--H--EE);

draw(A--EE,dashed);

draw(B--F);

draw(C--G);

draw(D--H);

label("$A$",A,S);

label("$B$",B,W);

label("$C$",C,S);

label("$D$",D,E);

label("$E$",EE,N);

label("$F$",F,W);

label("$G$",G,SW);

label("$H$",H,E);

[/asy]
Each side of $\triangle HAC$ is a face diagonal of the cube:

[asy]

import three;

triple A,B,C,D,EE,F,G,H;

A = (0,0,0);

B = (1,0,0);

C = (1,1,0);

D= (0,1,0);

EE = (0,0,1);

F = B+EE;

G = C + EE;

H = D + EE;

draw(B--C--D);

draw(B--A--D,dashed);

draw(EE--F--G--H--EE);

draw(A--EE,dashed);

draw(H--A--C,dashed);

draw(B--F);

draw(C--G);

draw(D--H--C);

label("$A$",A,NW);

label("$B$",B,W);

label("$C$",C,S);

label("$D$",D,E);

label("$E$",EE,N);

label("$F$",F,W);

label("$G$",G,SW);

label("$H$",H,E);

[/asy]

Therefore, $\triangle HAC$ is equilateral, so $\sin \angle HAC = \sin 60^\circ = \boxed{\frac{\sqrt{3}}{2}}$.